Linear Algebra: Subspace Proof

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To prove that the intersection of any collection of subspaces of V is a subspace of V, it is necessary to demonstrate two key properties: closure under addition and closure under scalar multiplication. When elements u and v are in the intersection, they must also belong to each subspace W within that intersection, ensuring that u + v is also in W due to the properties of subspaces. Similarly, for any scalar c and element u in the intersection, cu remains in W, confirming closure under scalar multiplication. This reasoning validates that the intersection itself satisfies the conditions to be a subspace. The discussion confirms the correctness of this approach to the proof.
*melinda*
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Prove that the intersection of any collection of subspaces of V is a subspace of V.

Ok, I know I need to show that:

1. For all u and v in the intersection, it must imply that u+v is in the intersection, and

2. For all u in the intersection and c in some field, cu must be in the intersection.

I can show both 1. and 2. for the trivial case where the intersection is zero, but I'm not sure what I need to do for the arbitrary case.

Any suggestions?
 
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Well, I'm not sure where you're stuck, so I'll throw out two random hints.


(a) Let W be one of the spaces in the intersection...

(b) What is the definition of "x is in the intersection"?
 
Since u and v are elements of the intersection, u and v will also be elements of any subspace W that is in the intersection. And since u and v are in W and W is a subspace, this guarantees that u+v will also be in W. This same argument would apply to scalar multiplication.

Is that the right idea?
 
Yes, that is one of the essential points of the proof.
 
Thanks!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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